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For which values of L does the vector [l,3,5] belong to span {[1,0,-2],[-3,1,7]?

Is thi set of vectors, $(2, 1), (3, 2), (1, 2)$, is linearly dependent or independent?Determine the Vectors Which Span R^4Basis and dimension of a subspace of $mathcalM_2times 2(mathbbR)$Understanding basis algorithm resultDetermining linear independence of a set of functionsWhich of the following functions lie in the span of the vector-valued functionsFind a basis for $V$ for which it is the dual basisAre these vectors linearly independent? 3 vectors.Determine whether the sets spans in $R^2$How can I tell if a matrix is singular or nonsingular?

0

$begingroup$

I'm not sure how to solve this problem. So far I have created the system of equations

$c_1 -3c_2 = l$;
$c_2 = 3$;
$-2c_1 + 7c_2 = -5$.

However, I cannot figure out how to solve the set of equations without knowing $l$.

linear-algebra matrices vectors

share|cite|improve this question

edited Mar 11 at 20:46

Subhasis Biswas

512411

asked Mar 11 at 20:25

Evan WilliamsEvan Williams

1

New contributor

Evan Williams is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Replace $c_2=3$ in the third equation.
  $endgroup$
  – hamam_Abdallah
  Mar 11 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Substitute $1 c_2 = 3$ into the last equation to figure are $c_1$. Then substitute both into the first equation to obtain $ell$.
  $endgroup$
  – Viktor Glombik
  Mar 11 at 20:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorGlombik duh, No idea how I missed that. Thanks a lot, everyone.
  $endgroup$
  – Evan Williams
  Mar 11 at 20:43
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm not sure how to solve this problem. So far I have created the system of equations

$c_1 -3c_2 = l$;
$c_2 = 3$;
$-2c_1 + 7c_2 = -5$.

However, I cannot figure out how to solve the set of equations without knowing $l$.

linear-algebra matrices vectors

share|cite|improve this question

edited Mar 11 at 20:46

Subhasis Biswas

512411

asked Mar 11 at 20:25

Evan WilliamsEvan Williams

1

New contributor

Evan Williams is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Replace $c_2=3$ in the third equation.
  $endgroup$
  – hamam_Abdallah
  Mar 11 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Substitute $1 c_2 = 3$ into the last equation to figure are $c_1$. Then substitute both into the first equation to obtain $ell$.
  $endgroup$
  – Viktor Glombik
  Mar 11 at 20:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ViktorGlombik duh, No idea how I missed that. Thanks a lot, everyone.
  $endgroup$
  – Evan Williams
  Mar 11 at 20:43
  

  
  
  
  

add a comment | 

$begingroup$

I'm not sure how to solve this problem. So far I have created the system of equations

$c_1 -3c_2 = l$;
$c_2 = 3$;
$-2c_1 + 7c_2 = -5$.

However, I cannot figure out how to solve the set of equations without knowing $l$.

linear-algebra matrices vectors

share|cite|improve this question

edited Mar 11 at 20:46

Subhasis Biswas

512411

asked Mar 11 at 20:25

Evan WilliamsEvan Williams

1

New contributor

Evan Williams is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 11 at 20:46

Subhasis Biswas

512411

asked Mar 11 at 20:25

Evan WilliamsEvan Williams

1

New contributor

Evan Williams is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 11 at 20:46

Subhasis Biswas

512411

asked Mar 11 at 20:25

Evan WilliamsEvan Williams

1

New contributor

Evan Williams is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 11 at 20:46

Subhasis Biswas

512411

edited Mar 11 at 20:46

Subhasis Biswas

512411

asked Mar 11 at 20:25

Evan WilliamsEvan Williams

1

New contributor

Evan Williams is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 20:25

Evan WilliamsEvan Williams

1

1 Answer
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0

$begingroup$

To find the values of L in which the vector $[L, 3, 5]$ is in the span of the vectors $[1, 0, -2]$ and $[-3, 1, 7]$, you must first ask yourself what it means for a vector to be a member of a spanning set.

In general, let $c_1, c_2, ... , c_k$ be a set of vectors. Then $Span$$c_1, c_2, ... , c_k$ is just the set of every possible linear combination of those vectors. In other words, $Span$$c_1, c_2, ... , c_k$ = $λ_1c_1 + λ_2c_2 + ... + λ_kc_k : λ_1, λ_2, ..., λ_k ∈ℝ$.

Now in our case, we are talking about $Span$$[1, 0, -,2], [-3, 1, 7]$. So if the vector $[L, 3, 5]$ is in our spanning set, it must mean that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ for some real numbers x and y. If you ignore the first columns of each vector, you can see that $[3, 5] = x[0, -2] + y[1, 7]$.

Note that y = 3. Thus, it must be that 5 = -2x + 7y implies that 5 = -2x + 7(3). Solve for x, and you get x = 8.

Looking back at our original vector equation, we see that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ must imply that $[L, 3, 5] = 8[1, 0, -2] + 3[-3, 1, 7]$ since we know our values of x and y

Finally, looking at the first column only, we see that L = 8(1) + 3(-3) = -1.

share|cite|improve this answer

edited Mar 12 at 5:33

answered Mar 11 at 21:13

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's something wrong with the last sentence of your second paragraph ! Typo in your definition of span.
  $endgroup$
  – Digitalis
  Mar 11 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Those should be plus signs, not commas.
  $endgroup$
  – amd
  Mar 12 at 5:31
  

  
  
  
  

add a comment | 

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0

$begingroup$

To find the values of L in which the vector $[L, 3, 5]$ is in the span of the vectors $[1, 0, -2]$ and $[-3, 1, 7]$, you must first ask yourself what it means for a vector to be a member of a spanning set.

In general, let $c_1, c_2, ... , c_k$ be a set of vectors. Then $Span$$c_1, c_2, ... , c_k$ is just the set of every possible linear combination of those vectors. In other words, $Span$$c_1, c_2, ... , c_k$ = $λ_1c_1 + λ_2c_2 + ... + λ_kc_k : λ_1, λ_2, ..., λ_k ∈ℝ$.

Now in our case, we are talking about $Span$$[1, 0, -,2], [-3, 1, 7]$. So if the vector $[L, 3, 5]$ is in our spanning set, it must mean that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ for some real numbers x and y. If you ignore the first columns of each vector, you can see that $[3, 5] = x[0, -2] + y[1, 7]$.

Note that y = 3. Thus, it must be that 5 = -2x + 7y implies that 5 = -2x + 7(3). Solve for x, and you get x = 8.

Looking back at our original vector equation, we see that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ must imply that $[L, 3, 5] = 8[1, 0, -2] + 3[-3, 1, 7]$ since we know our values of x and y

Finally, looking at the first column only, we see that L = 8(1) + 3(-3) = -1.

share|cite|improve this answer

edited Mar 12 at 5:33

answered Mar 11 at 21:13

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's something wrong with the last sentence of your second paragraph ! Typo in your definition of span.
  $endgroup$
  – Digitalis
  Mar 11 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Those should be plus signs, not commas.
  $endgroup$
  – amd
  Mar 12 at 5:31
  

  
  
  
  

add a comment | 

0

$begingroup$

To find the values of L in which the vector $[L, 3, 5]$ is in the span of the vectors $[1, 0, -2]$ and $[-3, 1, 7]$, you must first ask yourself what it means for a vector to be a member of a spanning set.

In general, let $c_1, c_2, ... , c_k$ be a set of vectors. Then $Span$$c_1, c_2, ... , c_k$ is just the set of every possible linear combination of those vectors. In other words, $Span$$c_1, c_2, ... , c_k$ = $λ_1c_1 + λ_2c_2 + ... + λ_kc_k : λ_1, λ_2, ..., λ_k ∈ℝ$.

Now in our case, we are talking about $Span$$[1, 0, -,2], [-3, 1, 7]$. So if the vector $[L, 3, 5]$ is in our spanning set, it must mean that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ for some real numbers x and y. If you ignore the first columns of each vector, you can see that $[3, 5] = x[0, -2] + y[1, 7]$.

Note that y = 3. Thus, it must be that 5 = -2x + 7y implies that 5 = -2x + 7(3). Solve for x, and you get x = 8.

Looking back at our original vector equation, we see that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ must imply that $[L, 3, 5] = 8[1, 0, -2] + 3[-3, 1, 7]$ since we know our values of x and y

Finally, looking at the first column only, we see that L = 8(1) + 3(-3) = -1.

share|cite|improve this answer

edited Mar 12 at 5:33

answered Mar 11 at 21:13

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  There's something wrong with the last sentence of your second paragraph ! Typo in your definition of span.
  $endgroup$
  – Digitalis
  Mar 11 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Those should be plus signs, not commas.
  $endgroup$
  – amd
  Mar 12 at 5:31
  

  
  
  
  

add a comment | 

$begingroup$

To find the values of L in which the vector $[L, 3, 5]$ is in the span of the vectors $[1, 0, -2]$ and $[-3, 1, 7]$, you must first ask yourself what it means for a vector to be a member of a spanning set.

In general, let $c_1, c_2, ... , c_k$ be a set of vectors. Then $Span$$c_1, c_2, ... , c_k$ is just the set of every possible linear combination of those vectors. In other words, $Span$$c_1, c_2, ... , c_k$ = $λ_1c_1 + λ_2c_2 + ... + λ_kc_k : λ_1, λ_2, ..., λ_k ∈ℝ$.

Now in our case, we are talking about $Span$$[1, 0, -,2], [-3, 1, 7]$. So if the vector $[L, 3, 5]$ is in our spanning set, it must mean that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ for some real numbers x and y. If you ignore the first columns of each vector, you can see that $[3, 5] = x[0, -2] + y[1, 7]$.

Note that y = 3. Thus, it must be that 5 = -2x + 7y implies that 5 = -2x + 7(3). Solve for x, and you get x = 8.

Looking back at our original vector equation, we see that $[L, 3, 5] = x[1, 0, -2] + y[-3, 1, 7]$ must imply that $[L, 3, 5] = 8[1, 0, -2] + 3[-3, 1, 7]$ since we know our values of x and y

Finally, looking at the first column only, we see that L = 8(1) + 3(-3) = -1.

share|cite|improve this answer

edited Mar 12 at 5:33

answered Mar 11 at 21:13

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

share|cite|improve this answer

edited Mar 12 at 5:33

answered Mar 11 at 21:13

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 11 at 21:13

TimTim

465

New contributor

Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 11 at 21:13

TimTim

465